You may alo use the partial fraction decomposition, such that:

`int 1/(x^2 - 4)dx = int 1/((x - 2)(x +2))dx`

`1/((x - 2)(x + 2)) = A/(x - 2) + B/(x + 2)`

`1 = A(x + 2) + B(x - 2)`

`1 = Ax + 2A + Bx - 2B => {(A+B = 0),(2A - 2B = 1):} => {(A = -B),(2A - 2B = 1):}`

`-2B - 2B = 1 => -4B = 1 => B = -1/4 => A = 1/4`

`1/((x - 2)(x + 2)) = (1/4)(1/(x - 2) - 1/(x + 2))`

`int 1/((x - 2)(x +2))dx = int(1/4)(1/(x - 2) - 1/(x + 2)) dx`

Using the property of linearity of integral yields:

`int 1/((x - 2)(x +2))dx = (1/4)int 1/(x - 2) dx -(1/4)int1/(x + 2) dx`

`int 1/((x - 2)(x +2))dx = (1/4) ln|x - 2| - (1/4) ln |x + 2| + c`

`int 1/((x - 2)(x +2))dx = (1/4)(ln|x - 2| - ln |x + 2|) + c`

Using the logarithmic identities yields:

`int 1/((x - 2)(x +2))dx = (1/4)(ln(|x - 2|/|x + 2|)) + c`

**Hence, evaluating the integral yields **`int 1/(x^2 - 4)dx = (1/4)(ln(|x - 2|/|x + 2|)) + c.`

We'll apply the formula of integrating a function:

f(x) = 1/(x^2 - a^2)

Int f(x)dx = Int dx/(x^2 - a^2)

Int dx/(x^2 - a^2) = (1/2a)*ln |(x-a)/(x+a)| + C

We recognize the denominator of the given function as being a difference of squares:

f(x) = 1/(x^2 - 4)

f(x) = 1/(x^2 - 2^2)

Int f(x)dx = Int dx/(x^2 - 2^2)

We'll substitute a by 2 and we'll get:

Int dx/(x^2 - 2^2) = (1/2*2)*ln |(x-2)/(x+2)| + C

Int dx/(x^2 - 2^2) = ln |(x-2)/(x+2)|^(1/4) + C